I don't understand the part that begins with "I differentiate the above equation ...".  You differentiate with respect to what?  And what does that have to do with a table?

@Markiyan Hirnyk Thanks for the link.  O'Rourke's theorem (page 126 in your referenced PDF) says that we need no more than floor( (n + 2h))/3 ) cameras, where n is the number of the vertices (counting vertices on the outer walls as well as on the holes) and h is the number of holes.

In the current problem we have 8 vertices on the outer walls and 4 vertices on the pollar, therefore n = 12.  We have one hole, therefore h = 1.  Then according to O'Rourke's theorem, the upper bound on the number of cameras is floor(14/3) = 4.  The solution that I posted earler, shows that 3 cameras will do, that is, the theorem's upper bound is an overestimate in this case.

@Markiyan Hirnyk Yes, I have.  I didn't see anything there pertaining to non-simple polygons.  Did you?

@vv The art gallery problem and the associated theorem by Chvatal/Fisk pertain to simple polygons, that is, polygons with no holes.  The "gallery" in the question asked here is not a simple polygon due to the pillar in the center.  There may exist extentions of the Chvatal/Fisk theorem to non-simple polygons but that is beyond my realm of knowledge.

@Markiyan Hirnyk That's quite curious and makes me wonder about the nature of this bug.  Your workaround is excellent when it works, but has its limitations.  The following still causes problems:

plots:-shadebetween(1/sqrt(1-x^2), 1/sqrt(1-x^3), x = 0 .. (1/2)*sqrt(3));

Two of the methods suggested on this page work and produce the correct result in this case:

1. The method suggested by acer
2. The method of "faking it" suggested by me, that is
   plots:-shadebetween(1/sqrt(1.0001-x^2), 1/sqrt(1.0001-x^3), x = 0 .. (1/2)*sqrt(3));
   
   

@DJJerome1976 Two ways to report bugs:

Method #1: Send mail to Support@maplesoft.com

Method #2: Look up the drop-down menu labeled "More" near the top of this page.  Within that menu select "Submit Software Change Request".

@darkspin1 I assume that when you wrote

v_1'(0)=v_2'(0)=v_1'(1)=v_2'(1)

you really meant

v_1'(0)=v_2'(0)=v_1'(1)=v_2'(1)=0.

Anyway, is there a reason that you insist on a discontinuous solution?  Your problem has a perfectly fine continuous solution where v1, v2, and u are constants:

 u = B/(1+j*B),

v1 = B/(1+j*B),

v2 = 1/(1+j*B).

@darkspin1 You have

f(u) = u+1 for u < 0 and u-1 for u>0

and

-v + j*u + f(u) = 0,

that is,

j*u + f(u) = v.

You claim that we can solve this equation for u in terms of v.  That is not correct.  If you plot the graph of v =  j*u + f(u) versus u, you will see that any horizontal line v = c  with -1 < c < +1 intersects the graph in two points.   Therefore there is no unique u corresponding to such a v.

I have not examined the rest of your problem, but the point noted above can explain why your solution is not what you expect it to be.

Edgardo, I did not mean anything negative with my comments.   On the contrary, I appreciate your immediate response and solution to the issues raised in this thread.  I tested your Maple16_DE_update.mla.zip with Maple 2015 and it works as advertized.  Thanks!

@darkspin1 Be careful there!  Maple 2015 fails on the homogeneous equation as well:

de := diff(x(t), t$4)+4*diff(x(t), t$2)-4*x(t);

dsol := dsolve(de);

simplify(dsol, size);

We see that the multipliers of _C2 and _C4 are the same!

@Markiyan Hirnyk Although minimize() is the easiest method to use, it does not always work.  Try, for instance, changing the population of A from 5 to 8:

minimize(8*abs(x-3)+8*abs(y-7)+2*abs(x-6)+2*abs(y-5)+2*abs(x-5)+2*abs(y-3)+abs(x-1)+abs(y-2)+3*abs(x-7)+3*abs(y-1), x = -infinity .. infinity, y = -infinity .. infinity, location);

     Error, (in minimize/allvalues) to be implemented

Consider the points P = [ a, b ] and Q = [ c, d].  How would you express the taxicab distance between P and Q in terms of a, b, c, d?

Some clarification can help:

Does "center" mean the vertex?

Does "eccentricity" mean the cone's vertex angle?

Are you asking for a right circular cone or a more general cone?

@Christopher2222 Here is what I get.  It looks good me.